题意：
There is a polyline going through points (0, 0) – (x, x) – (2x, 0) – (3x, x) – (4x, 0) – ... - (2kx, 0) – (2kx + x, x) – ....
We know that the polyline passes through the point (a, b). Find minimum positive value x such that it is true or determine that there is no such x.
Input
Only one line containing two positive integers a and b (1 ≤ a, b ≤ 109).
Output
Output the only line containing the answer. Your answer will be considered correct if its relative or absolute error doesn't exceed 10 - 9. If there is no such x then output - 1 as the answer.
思路：先大概确定范围，再二分。
代码：
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<queue>
#include<algorithm>
#include<cmath>
#include<map>
using namespace std;
#define INF 0x7fffffff
int main(){
int a,b,x,y;
double ans = INF;
while(scanf("%d%d",&x,&y) == 2){
if(y > x){
printf("-1\n");
continue;
}
else if(y == 0){
printf("0.000000000001\n");
continue;
}
b = x+y;
int k = 1;
int t ;
ans = INF ;
for(; ;){
t = k;
k *= 2;
if(b*1.0/k >= y)
ans = ans < b*1.0/k ? ans : b*1.0/k ;
else
break;
}
while(t<k){
int mid = (t+k)/2;
if(mid%2 == 1)
break;
if(b*1.0/mid >= y)
t = mid ;
else
k = mid ;
}
if(t == 1)
t = 2;
if(k == 1)
k = 2;
double s1 = b*1.0/k;
double s2 = b*1.0/t;
if(s1 >= y && s2 >= y)
ans = s1 > s2 ? s2 : s1 ;
else if(s1 >= y)
ans = s1;
else
ans = s2;
b = y -x ;
if(b<0)
b = -b;
if(b == 0){
printf("%.12lf\n",(double)y);
continue;
}
k = 1;
for(; ;){
t = k;
k *= 2;
if(b*1.0/k >= y)
ans = ans < b*1.0/k ? ans : b*1.0/k ;
else
break;
}
while(t<k){
int mid = (t+k)/2;
if(mid%2 == 1)
break;
if(b*1.0/mid >= y)
t = mid ;
else
k = mid ;
}
if(t == 1)
t = 2;
if(k == 1)
k = 2;
s1 = b*1.0/k;
s2 = b*1.0/t;
if(s1 >= y && s2 >= y){
double temp = s1 > s2 ? s2 : s1 ;
ans = temp;
}
else if(s1 >= y)
ans = ans > s1 ? s1 : ans;
else if(s2 >= y)
ans = ans > s2 ? s2 : ans;
printf("%.12lf\n",ans);
}
return 0;
}